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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p>Further, one has the system governing the displacement as</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_19.html">
\begin{equation}
\begin{aligned}
&amp;\frac{\textrm{d} x}{\textrm{d} t}=v=\left(v_0+\frac{mg}{k} \right) e^{-\frac{k}{m} t}-\frac{mg}{k},\\
&amp;x(0)=0.
\end{aligned}\tag{2.5.3}
\end{equation}
</div>
<p class="continuation">From (<a href="" class="xref" data-knowl="./knowl/eq2_19.html" title="Equation 2.5.3">(2.5.3)</a>), one has</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_19.html">
\begin{equation*}
x=-\frac{m}{k} \left(v_0+\frac{mg}{k}\right) e^{-\frac{k}{m}t}-\frac{mg}{k}t+D.
\end{equation*}
</div>
<p class="continuation">The initial condition gives</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_19.html">
\begin{equation*}
D=\frac{m}{k} \left(v_0+\frac{mg}{k}\right).
\end{equation*}
</div>
<p class="continuation">Thus, the solution to the displacement is</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_19.html">
\begin{equation}
x=\frac{m}{k} \left( v_0+\frac{mg}{k} \right) \left( 1-e^{-\frac{k}{m}t}  \right)-\frac{mg}{k}t.\tag{2.5.4}
\end{equation}
</div>
<span class="incontext"><a href="sec2_5.html#p-37" class="internal">in-context</a></span>
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